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Re: large integration result for simple problem: 1/x,, also BesselJ

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122859] Re: large integration result for simple problem: 1/x,, also BesselJ
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 12 Nov 2011 07:36:44 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201111110955.EAA08514@smc.vnet.net> <7999FCC4-7E77-4ED4-AEF6-AF3CCFAD0FA7@mimuw.edu.pl> <4EBD4CAB.1090505@eecs.berkeley.edu>

Mathematica 8 returns:


Integrate[BesselJ[n, b*x], {x, 0, Infinity},
 Assumptions -> {Re[n] > -1}]

 b^(n - 2)*(b^2)^(1/2 - n/2)

Andrzej Kozlowski


On 11 Nov 2011, at 17:26, Richard Fateman wrote:

> On 11/11/2011 7:46 AM, Andrzej Kozlowski wrote:
>> Mathematica 8 gives:
>>
>> Integrate[BesselJ[n, b*x], {x, 0, Infinity},  Assumptions ->  {Re[n]> -1, Im[b] == 0}]
>>
>> Sign[b]^n/Abs[b]
> Mathematica 7 gives the same answer with those assumptions.
>
> What's your point?  In my command, where I leave off the Im[b]==0 , I got an answer that includes..
>
> If  [  Abs[Im[b]] == 0, b^(-2 + n) (b^2)^(1/2 - n/2), .....
>
>
> To me,  Abs[Im[b]] == 0  means the same as Im[b]==0, so why does the answer say Abs[...]==0?
> Why is the user required to feed back into the Integrate program the condition it derives, as an Assumption,
> in order to get a simplified result?
>
> Does Mathematica 8 return  b^(-2 + n) (b^2)^(1/2 - n/2)  ?
>
>
> Does Mathematica 8 do any better for integrating 1/x ?  (Sorry, UC Berkeley seems to have let their upgrade lapse).
> Wolfram Alpha  times out on this.
> It also times out on the Bessel integral.
>
> RJF
>
>
>
>
>
>>
>>
>>
>> Note the additional assumption on b. Without it the result is clearly not true.
>>
>> Andrzej Kozlowski
>>
>>
>> On 11 Nov 2011, at 10:55, Richard Fateman wrote:
>>
>>> (at least in Mathematica 7.0)
>>>
>>> try
>>>
>>> Integrate[1/x,{x,a,b}]
>>>
>>> Also
>>>
>>> Integrate[BesselJ[n, b*x], {x, 0, Infinity},   Assumptions ->  Re[n]>  -1 ]
>>>
>>> which returns an expression including this ....
>>>
>>> b^(-2 + n) (b^2)^(1/2 - n/2)
>>>
>>> which should be possible to simplify.  For example, for b>0, the
>>> expression is 1/b.
>>>
>>> maybe  1/b * If[b>0, 1 ,  -(-1)^n]  or so.
>>>
>>> I've been playing with integration of expressions involving Bessel
>>> functions.  Mathematica is sometimes surprising, on both sides of the
>>> ledger -- (Yes we Can and No we Can't).
>>>
>>>
>>> RJF
>>>
>>>
>




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